Convolution & LTI systems worksheet with answers

Free printable Convolution & LTI systems worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
x[n]n−2−1202132h[n]n−10−21−22−13

The sequence x[n] is nonzero for n = -1…1 and h[n] for n = 0…2, as shown. Find the last index nlast at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = -1…1h[n]: n = 0…2
nlast =
2.Easy
x[n]n−10123−2445−1617h[n]n−10111223

The sequence x[n] is nonzero for n = 3…6 and h[n] for n = 0…2, as shown. Find the first index nfirst at which y[n] = x[n] ∗ h[n] can be nonzero.

x[n]: n = 3…6h[n]: n = 0…2
nfirst =
3.Medium
x(t)t21h(t)t1

The pulse x(t) = 1 for 0 ≤ t < 2 s (zero elsewhere) is applied to a system with h(t) = e−0.5tu(t). Find the output y(2.5) at t = 2.5 s, after the pulse has ended.

x(t) = 1, 0 ≤ t < 2 sh(t) = e−0.5tu(t)t = 2.5 s
y(2.5) =
4.Medium
x(t)t25h(t)t21

The pulse x(t) = 5 for 0 ≤ t < 2 s is convolved with h(t), a triangle that falls linearly from 1 at t = 0 to 0 at t = 2 s. Find the area Ay under y(t) = x(t) ∗ h(t).

x(t) = 5, 0 ≤ t < 2 sh(0) = 1, h(2) = 0
Ay =
5.Hard
10cos(7.5t)h(t) = 2e−5t​u(t)y(t)

The input x(t) = 10·cos(7.5t) is applied to a system with h(t) = 2·e−5tu(t). Find the amplitude Ym of the steady-state output.

h(t) = 2e−5tu(t)x(t) = 10cos(7.5t)
Ym =
6.Hard
δ(t)y′ + 10y = 3xh(t)

A system obeys y′(t) + 10·y(t) = 3·x(t). Find its impulse response at t = 0.2 s, h(0.2).

y′ + 10y = 3xt = 0.2 s
h(0.2) =

Answer key

  1. 1. nlast = 3
  2. 2. nfirst = 3
  3. 3. y(2.5) = 0.9846
  4. 4. Ay = 10
  5. 5. Ym = 2.219
  6. 6. h(0.2) = 0.406

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